# Regularity results and asymptotic behavior for a noncoercive parabolic problem

## Abstract

In this paper we study the regularity and the behavior in time of the solutions to a quasilinear class of noncoercive problems whose prototype is

\begin{aligned} \left\{ \begin{array}{ll} u_{t} - \mathrm{div}(a(x,t,u)\nabla u) = -\mathrm{div}(u\,E(x,t)) &{}\quad \text{ in }\, \Omega \times (0,T), \\ u(x,t) = 0 &{}\quad \text{ on }\, \partial \Omega \times (0,T), \\ u(x,0) = u_{0}(x) &{}\quad \text{ in }\, \Omega . \end{array} \right. \end{aligned}

In particular we show that under suitable conditions on the vector field E, even if the problem is noncoercive and although the initial datum $$u_0$$ is only an $$L^{1}(\Omega )$$ function, there exist solutions that immediately improve their regularity and belong to every Lebesgue space. We also prove that solutions may become immediately bounded. Finally, we study the behavior in time of such regular solutions and we prove estimates that allow to describe their blow-up for t near zero.

## Introduction

Let us consider the following parabolic problem

\begin{aligned} \left\{ \begin{array}{ll} u_{t}(x,t) - \mathrm{div}(a(x,t,u(x,t))\nabla u(x,t)) = -\mathrm{div}(u(x,t)\,E(x,t)) &{}\quad \text{ in }\, Q_{T} \equiv \Omega \times (0,T), \\ u(x,t) = 0 &{}\quad \text{ on }\, \partial \Omega \times (0,T), \\ u(x,0) = u_{0}(x) &{} \quad \text{ in }\, \Omega , \end{array} \right. \end{aligned}
(1.1)

where $$\Omega$$ is an open bounded subset of $${\mathbb {R}}^N$$, $$N > 2$$, and $$T>0$$. We assume that $$a(x,t,s) : Q_{T} \times {\mathbb {R}}\rightarrow {\mathbb {R}}$$ is a Carathéodory function (i.e., measurable in (xt) for every s in $${\mathbb {R}}$$, and continuous in s for almost every (xt) in $$Q_{T}$$) such that

\begin{aligned} 0 < \alpha \le a(x,t,s) \le \beta , \end{aligned}
(1.2)

for almost every (xt) in $$Q_{T}$$ and for every s in $${\mathbb {R}}$$, where $$\alpha$$ and $$\beta$$ are positive constants. On the initial datum we assume

\begin{aligned} u_{0} \in L^{1}(\Omega ). \end{aligned}
(1.3)

The main difficulty in studying this problem is the presence of the lower order term $$-\mathrm{div}(u\,E(x,t))$$ which makes the problem noncoercive. In addition, the initial datum has a very low regularity, so that the study of such an equation is more complicated.

In the stationary case this problem was studied in the sixties by Stampacchia (see [14, 15]) when E “is not too large”, and then by several authors until nowadays (see for example [2, 3, 6, 8] and the references therein).

In the evolution case existence and regularity results can be found in  (see also ).

We recall that if E is zero then a very surprising phenomenon appears: even if $$u_0$$ is only a summable function, there exists a solution u of (1.1) that becomes immediately (i.e., for every $$t > 0$$) bounded and satisfies the same decay (or ultracontractive estimate) of the solution of the heat equation, i.e.,

\begin{aligned} \Vert u(t)\Vert _{_{L^{\infty }(\Omega )}} \le C(N,\alpha ) \, \frac{\Vert u_{0}\Vert _{_{L^{1}(\Omega )}}}{t^{\frac{N}{2}}} \quad \mathrm{for}\,\mathrm{every} \,\, t \in (0,T), \end{aligned}
(1.4)

(see [5, 12, 16]).

The aim of this paper is to understand what happens when E is a nonzero measurable vector field; in particular, we want to understand if an immediate improvement in the regularity of the solutions appears or not. We will prove (see Sect. 2) that if E satisfies

\begin{aligned} |E(x,t)| \le \gamma \,|B(x)| \qquad B \in L^{N}(\Omega ), \end{aligned}
(1.5)

for almost every (xt) in $$Q_{T}$$, then there is an immediate regularization since there exists a solution u of (1.1) which belongs to $$L^\infty (t_0,T;L^{q}(\Omega ))$$ for every $$1< q <+\infty$$ and $$t_0$$ in (0, T). Besides, we will derive quantitative estimates “near $$t=0$$” (see Sect. 2.1) and for t large (see Sect. 2.2).

Moreover, we show that under stronger assumptions on E solutions become immediately bounded (see Sect. 3); furthermore, near “$$t = 0$$” the same behavior (1.4) of the solution of the heat equation holds true, but in this case the constant c in (1.4) depends also on E.

Finally, in Sect. 4 we study the less regular case of a vector field E such that

\begin{aligned} |E(x,t)| \le \frac{\gamma }{|x|}, \quad \quad \gamma >0, \end{aligned}

which may not belong to $$L^{\infty }(0,T;L^{N}(\Omega ))$$ if 0 belongs to $$\Omega$$.

### Remark 1.1

We would like to thank the referee of the paper for pointing out that in the autonomous case, i.e., for the equation

\begin{aligned} u_{t} - \mathrm{div}(A(x)\nabla u) = -\mathrm{div}(u\,E(x)), \end{aligned}
(1.6)

with E in $$L^{\infty }(\Omega )$$, the following holds:

• the operator

\begin{aligned} A(u) = - \mathrm{div}(A(x)\nabla u) + \mathrm{div}(u\,E(x)) \end{aligned}

is quasi accretive in $$L^{q}(\Omega )$$ spaces (for every $$1 \le q \le \infty$$), and quasi m-accretive in $$L^{q}(\Omega )$$ spaces (for every $$1 \le q < \infty$$);

• every mild solution u of (1.6) satisfies an $$L^{1}$$$$L^{\infty }$$ estimate similar to (1.4) thanks to Theorem 1.2 of .

## An improvement of regularity

In this section we will show that even if the initial datum $$u_0$$ is only assumed to be a summable function, if E satisfies (1.5) there exists a solution of (1.1) which belongs to every Lebesgue space $$L^{\infty }(t_0,T;L^{q}(\Omega ))$$ for every $$1< q < +\infty$$ and $$t_0 \in (0,T)$$ (see Theorem 2.1 in Sect. 2.1 below). Moreover, we describe the blow-up of the $$L^{q}(\Omega )$$-norm of u(t) as t tends to zero and the behavior of the solution for t large.

We recall that here by a solution of (1.1) we mean a function u in $$L^{\infty }(0,T; L^{1}(\Omega ))$$ and $$L^{1}(0,T;W^{1,1}_{0}(\Omega ))$$ such that

\begin{aligned} \int _{Q_{T}}\,[-u\,\varphi _{t} + a(x,t,u)\nabla u \nabla \varphi ] = \int _{Q_{T}}\,u\,E\,\nabla \varphi + \int _{\Omega }\,u_{0}\,\varphi (0), \end{aligned}

for every $$\varphi$$ in $$W^{1,\infty }(0,T;L^{\infty }(\Omega )) \cap L^{\infty }(0,T;W^{1,\infty }_{0}(\Omega ))$$ such that $$\varphi (T) = 0$$.

We recall that, thanks to the results of , under assumptions (1.3) and (1.5) there exists a solution u of (1.1), with u in $$L^{\infty }(0,T;L^{1}(\Omega )) \cap L^{q}(0,T;W^{1,q}_{0}(\Omega ))$$, with $$q = \frac{N+2}{N+1}$$ (see Lemma 3.2 of ). Furthermore, if |E| belongs to $$L^{\infty }(Q_{T})$$ and $$u_{0}$$ belongs to $$L^{\infty }(\Omega )$$, there exists a bounded solution u of (1.1) such that (see Lemma 3.1 of )

\begin{aligned} \Vert u(t)\Vert _{_{L^{1}(\Omega )}} \le \Vert u_{0}\Vert _{_{L^{1}(\Omega )}} \qquad \text { for } \text {almost } \text { every }\, t \in (0,T). \end{aligned}
(2.1)

### Theorem 2.1

Assume (1.2), (1.3) and (1.5). Then there exists a solution u of (1.1) satisfying

\begin{aligned} u \in L^{\infty }(t_0,T;L^{q}(\Omega )) \quad \text{ for } \text{ every }\, t_{0}\, \text{ in }\, (0,T),\, \text{ and } \text{ for } \text{ every }\, 1< q < +\infty . \end{aligned}
(2.2)

Furthermore, the following estimate holds:

\begin{aligned} \Vert u\Vert _{_{L^{\infty }(t_0,T;L^{q}(\Omega ))}} \le c_0\, \Vert u_{0}\Vert _{_{L^{1}(\Omega )}} \end{aligned}
(2.3)

where $$c_0$$ is a constant depending only on $$t_0$$, q, N, $$\alpha$$, T and B. Moreover, for every $$q \ge 2$$ and $$t_0 >0$$ it results

\begin{aligned} |\nabla |u|^{\frac{q}{2}-1}u| \in L^2(t_0,T;L^{2}(\Omega )), \quad |\nabla u| \in L^2(t_0,T;L^{2}(\Omega )). \end{aligned}

Finally,

\begin{aligned} \Vert u(t)\Vert _{_{L^{q}(\Omega )}} \le c_1 \,\frac{\Vert u_{0}\Vert _{_{L^{1}(\Omega )}}}{t^{\frac{N}{2}\left( 1-\frac{1}{q}\right) }} \qquad \text { for } \text { almost } \text { every }\, t\, \text { in }\, (0,T), \end{aligned}
(2.4)

where $$c_1$$ is a constant depending only on N, q, $$\alpha$$, $$\gamma$$, T and B (see formula (2.21)).

### Proof

In  it is proved the existence of a solution u of (1.1), obtained as the almost everywhere limit of a sequence of weak solutions $$u_{n}$$ in $$L^2(0,T;W^{1,2}_{0}(\Omega )) \cap C([0,T];L^2(\Omega ))$$ of the following problems:

\begin{aligned} \left\{ \begin{array}{ll} (u_{n})_{t} - \mathrm{div}(a(x,t,u_{n})\nabla u_{n}) = -\mathrm{div}(T_n(u_{n})\,E_{n}(x,t)) &{}\quad \text{ in }\, Q_{T} \equiv \Omega \times (0,T), \\ u_{n}(x,t) = 0 &{}\quad \text{ on }\, \partial \Omega \times (0,T), \\ u_{n}(x,0) = T_n(u_{0}(x)) &{}\quad \text{ in }\, \Omega , \end{array} \right. \nonumber \\ \end{aligned}
(2.5)

where $$T_n$$ is the usual truncating function

\begin{aligned} T_n(s) = \min \{ n,\max \{ -n,s\}\}, \qquad s \in {\mathbb {R}}, \end{aligned}

and

\begin{aligned} E_{n}(x,t) = \frac{E(x,t)}{1 + \frac{1}{n}|E(x,t)|}. \end{aligned}
(2.6)

Notice that since $$T_n(u_{n})\,E_{n}(x,t)$$ belongs to $$L^{\infty }(Q_{T})$$ then $$u_{n}$$ also belongs $$L^{\infty }(Q_{T})$$ by the results of . We fix $$q > 1$$, $$t > 0$$, choose as test function in (2.5) $$v_n = |u_{n}|^{q-2}\,u_{n}$$, and integrate on $$\Omega$$. We obtain, using (1.2), and the fact that $$|T_{n}(s)| \le |s|$$,

\begin{aligned} \begin{array}{l} \displaystyle \frac{1}{q} \frac{d}{dt} \int _{\Omega }\,|u_{n}(t)|^{q} + \alpha \,(q-1) \int _{\Omega }|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} \\ \displaystyle \qquad \le (q-1) \int _{\Omega }T_n(u_{n}(t))\,E_{n}\cdot \nabla u_{n}(t) \, |u_{n}(t)|^{q-2}\,\\ \displaystyle \qquad \le (q-1) \int _{\Omega }|E|\, |\nabla u_{n}(t)|\,|u_{n}(t)|^{q-1}. \end{array} \end{aligned}
(2.7)

We now work with the right hand side; using Young inequality, as well as (1.5), we have

\begin{aligned} \begin{array}{rcl} \displaystyle \int _{\Omega }|E|\, |\nabla u_{n}(t)|\,|u_{n}(t)|^{q-1} &{}{} \le &{}{}\displaystyle \gamma \,\int _{\Omega }|\nabla u_{n}(t)|\,|B(x)|\,|u_{n}(t)|^{q-1} \\ &{}{} \le &{}{}\displaystyle \frac{\alpha }{2}\,\int _{\Omega }\,|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} + C_1\,\int _{\Omega }\,|B(x)|^{2}\,|u_{n}(t)|^{q},\end{array} \end{aligned}

where $$C_1=\frac{\gamma ^2}{2\alpha }$$. Therefore, part of the right hand side can be absorbed in the left hand one, to obtain

\begin{aligned} \frac{1}{q} \frac{d}{dt} \int _{\Omega }\,|u_{n}(t)|^{q} + \frac{\alpha \,(q-1)}{2} \int _{\Omega }|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} \le C_2\,\int _{\Omega }\,|B(x)|^{2}\,|u_{n}(t)|^{q},\nonumber \\ \end{aligned}
(2.8)

where $$C_2 = C_{2}(q) = C_1\,(q-1)$$. We continue to work on the right hand side; we have, for some $$\rho > 0$$, and thanks to Hölder inequality,

\begin{aligned}&\int _{\Omega }\,|B(x)|^{2}\,|u_{n}(t)|^{q} = \int _{\{|B(x)| \le \rho \}}\,|B(x)|^{2}\,|u_{n}(t)|^{q} + \int _{\{|B(x)|> \rho \}}\,|B(x)|^{2}\,|u_{n}(t)|^{q} \nonumber \\&\quad \le \rho ^{2}\,\int _{\Omega }\,|u_{n}(t)|^{q} + \int _{\{|B(x)|> \rho \}}\,|B(x)|^{2}\,|u_{n}(t)|^{q} \nonumber \\&\quad \le \rho ^{2}\,\int _{\Omega }\,|u_{n}(t)|^{q} + \Big ( \int _{\{|B(x)| > \rho \}}\,|B(x)|^{N}\Big )^{\frac{2}{N}} \Big ( \int _{\Omega }\, |u_{n}(t)|^{\frac{2^{*}}{2}q}\Big )^{\frac{2}{2^{*}}}, \end{aligned}
(2.9)

where, as usual, $$2^{*} = \frac{2N}{N-2}$$ is the Sobolev embedding exponent. Since, by Sobolev embedding, we have

\begin{aligned} \Big ( \int _{\Omega }\, |u_{n}(t)|^{2^{*}\frac{q}{2}}\Big )^{\frac{2}{2^{*}}} \le {\mathcal {S}}\,\int _{\Omega }\, |\nabla [|u_{n}(t)|^{\frac{q}{2}}\mathrm {sign}(u_{n})]|^{2} = \frac{q^{2}\,{\mathcal {S}}}{4}\,\int _{\Omega }\,|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2},\end{aligned}

we therefore have that

\begin{aligned}&C_2\,\int _{\Omega }\,|B(x)|^{2}\,|u_{n}(t)|^{q} \le C_2\rho ^{2}\,\int _{\Omega }\,|u_{n}(t)|^{q} \\&\quad + C_2\,\frac{q^{2}\,{\mathcal {S}}}{4}\Big ( \int _{\{|B(x)| > \rho \}}\,|B(x)|^{N}\Big )^{\frac{2}{N}} \, \int _{\Omega }\,|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2}. \end{aligned}

We now choose $$\rho$$ in such a way that

\begin{aligned} C_2\,\frac{q^{2}\,{\mathcal {S}}}{4}\Big ( \int _{\{|B(x)| > \rho \}}\,|B(x)|^{N}\Big )^{\frac{2}{N}} \le \frac{\alpha \,(q-1)}{4}. \end{aligned}
(2.10)

Such a choice is possible since B belongs to $$L^{N}(\Omega )$$. Note that $$\rho$$ does not depend on t. We therefore have that

\begin{aligned} C_2\,\int _{\Omega }\,|B(x)|^{2}\,|u_{n}(t)|^{q} \le C_3\,\int _{\Omega }\,|u_{n}(t)|^{q} + \frac{\alpha \,(q-1)}{4}\,\int _{\Omega }\,|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2},\nonumber \\ \end{aligned}
(2.11)

where $$C_3$$ is a constant depending only on $$\alpha$$, q, $$\gamma$$, $${\mathcal {S}}$$ and B. Substituting this inequality in (2.8) we obtain, after simplifying equal terms,

\begin{aligned} \frac{1}{q} \frac{d}{dt} \int _{\Omega }\,|u_{n}(t)|^{q} + \frac{\alpha \,(q-1)}{4} \int _{\Omega }\, |\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} \le C_3\,\int _{\Omega }\,|u_{n}(t)|^{q}. \end{aligned}
(2.12)

Writing $$|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} = \frac{4}{q^{2}}\,|\nabla [|u_{n}(t)|^{\frac{q}{2}}\mathrm{sign}(u_{n})]|^{2}$$, and using Sobolev embedding, we arrive at

\begin{aligned} \frac{1}{q} \frac{d}{dt} \int _{\Omega }\,|u_{n}(t)|^{q} + \frac{\alpha \,(q-1)}{q^{2}{\mathcal {S}}} \Big ( \int _{\Omega }\,|u_{n}(t)|^{\frac{2^{*}}{2}q}\Big )^{\frac{2}{2^{*}}} \le C_3\,\int _{\Omega }\,|u_{n}(t)|^{q}. \end{aligned}
(2.13)

Since $$1< q < \frac{2^{*}}{2}\,q$$, we can interpolate, to obtain

\begin{aligned} \int _{\Omega }\,|u_{n}(t)|^{q} \le \Big ( \int _{\Omega }\, |u_{n}(t)| \Big )^{q\,(1-\theta )} \, \Big ( \int _{\Omega }\,|u_{n}(t)|^{\frac{2^{*}}{2}q} \Big )^{\frac{2}{2^{*}}\,\theta }, \end{aligned}
(2.14)

where $$\theta$$ in (0, 1) is such that

\begin{aligned} \frac{1}{q} = (1 - \theta )\,\frac{1}{1} + \theta \,\frac{2}{2^{*}q}. \end{aligned}

We recall now (see the proof of Lemma 3.1 in ) that (2.1) holds true for $$u_{n}$$, so that

\begin{aligned} \Vert u_{n}(t)\Vert _{_{L^{1}(\Omega )}} \le \Vert T_{n}(u_{0})\Vert _{_{L^{1}(\Omega )}}, \qquad \text{ for } \text{ every }\, t\, \text{ in }\, (0,T). \end{aligned}

Hence, from (2.14) it follows that

\begin{aligned}&\int _{\Omega }\,|u_{n}(t)|^{q} \le \Big ( \int _{\Omega }\, |T_n(u_{0})| \Big )^{q\,(1-\theta )} \, \Big ( \int _{\Omega }\,|u_{n}(t)|^{\frac{2^{*}}{2}q} \Big )^{\frac{2}{2^{*}}\,\theta } \\&\quad \le \Big ( \int _{\Omega }\, |u_{0}| \Big )^{q\,(1-\theta )} \, \Big ( \int _{\Omega }\,|u_{n}(t)|^{\frac{2^{*}}{2}q} \Big )^{\frac{2}{2^{*}}\,\theta }. \end{aligned}

Therefore,

\begin{aligned} \Big ( \int _{\Omega }\,|u_{n}(t)|^{\frac{2^{*}}{2}q} \Big )^{\frac{2}{2^{*}}\,\theta } \ge \frac{1}{\Vert u_{0}\Vert _{_{L^{1}(\Omega )}}^{q\,(1-\theta )}}\int _{\Omega }\,|u_{n}(t)|^{q}, \end{aligned}
(2.15)

where we have assumed that $$u_0 \not \equiv 0$$ (otherwise there is nothing to prove). Inserting (2.15) into (2.13), we obtain

\begin{aligned} \frac{d}{dt} \int _{\Omega }\,|u_{n}(t)|^{q} + A\,\Big ( \int _{\Omega }\,|u_{n}(t)|^{q} \Big )^{\frac{1}{\theta }} \le C_3\,q\,\int _{\Omega }\,|u_{n}(t)|^{q}, \end{aligned}

where

\begin{aligned} A =\frac{\alpha \,(q-1)}{q {\mathcal {S}} \Vert u_0\Vert _{L^{1}(\Omega )}^{\frac{q(1-\theta )}{\theta }}}. \end{aligned}

Since

\begin{aligned} \frac{1}{\theta } = 1 + \delta , \qquad \text{ with } \qquad \delta = \frac{2}{N(q-1)}, \end{aligned}

we thus have

\begin{aligned} \frac{d}{dt} \int _{\Omega }\,|u_{n}(t)|^{q} + A\,\Big ( \int _{\Omega }\,|u_{n}(t)|^{q} \Big )^{1 + \delta } \le C\,\int _{\Omega }\,|u_{n}(t)|^{q}, \end{aligned}
(2.16)

where to simplify the notation we have set $$C=C_3\,q$$. Define now

\begin{aligned} y(t) = \int _{\Omega }\,|u_{n}(t)|^{q}. \end{aligned}

From (2.16) we have that y(t) is such that

\begin{aligned} y'(t) + A\,y(t)^{1 + \delta } \le C\,y(t), \end{aligned}

where A and C are as above. Multiplying by $$\mathrm{e}^{-C\,t}$$, we have

\begin{aligned} \mathrm{e}^{-C\,t}\,y'(t) - C\,\mathrm{e}^{-C\,t}y(t) + A\,\mathrm{e}^{-C\,t}y(t)^{1 + \delta } \le 0, \end{aligned}

that is,

\begin{aligned}{}[\mathrm{e}^{-C\,t}\,y(t)]' + A\,\mathrm{e}^{-C\,t}y(t)^{1 + \delta } \le 0. \end{aligned}

Define $$z(t) = \mathrm{e}^{-C\,t}\,y(t)$$, so that we have

\begin{aligned} z'(t) + A\,\mathrm{e}^{-C\,t}\,[\mathrm{e}^{C\,t}\,z(t)]^{1 + \delta } \le 0, \end{aligned}

that is

\begin{aligned} z'(t) + A\,\mathrm{e}^{C\,\delta \,t}\,z(t)^{1+\delta } \le 0. \end{aligned}

Dividing by $$z(t)^{1+\delta }$$, we have

\begin{aligned} \frac{z'(t)}{z(t)^{1+\delta }} + A\,\mathrm{e}^{C\,\delta \,t} \le 0, \end{aligned}

which can be rewritten as

\begin{aligned} \Big [ -\frac{1}{\delta \,z(t)^{\delta }} + \frac{A\,\mathrm{e}^{C\,\delta \,t}}{C\,\delta } \Big ]' \le 0, \end{aligned}

so that the function

\begin{aligned} t \mapsto \frac{A\,\mathrm{e}^{C\,\delta \,t}}{C\,\delta } - \frac{1}{\delta \,z(t)^{\delta }} \quad \text{ is } \text{ decreasing, } \end{aligned}

or, simplifying $$\delta$$, that the function

\begin{aligned} t \mapsto A\,\mathrm{e}^{C\,\delta \,t} - C\,z(t)^{-\delta } \quad \text{ is } \text{ decreasing, } \end{aligned}
(2.17)

Thanks to (2.17), we have that, for every $$0< t < T$$,

\begin{aligned} A\,\mathrm{e}^{C\,\delta \,t} - C\,z(t)^{-\delta } \le A - C\,z(0)^{-\delta } \le A , \end{aligned}

and thus,

\begin{aligned} z(t)^{-\delta } \ge A C^{-1}\,(\mathrm{e}^{C\,\delta \,t} - 1) \qquad \text{ for } \text{ every }\, t \in (0,T), \end{aligned}

that is,

\begin{aligned} z(t) \le \frac{1}{[A C^{-1}\,(\mathrm{e}^{C\,\delta \,t} - 1) ]^{\frac{1}{\delta }}} \qquad \text{ for } \text{ every }\, t \in (0,T), \end{aligned}

Recalling the definition of z(t), we thus have that

\begin{aligned} y(t) \le \frac{\mathrm{e}^{C\,t}}{[A C^{-1}\,(\mathrm{e}^{C\,\delta \,t} - 1) ]^{\frac{1}{\delta }}} \qquad \text{ for } \text{ every }\, t \in (0,T). \end{aligned}

that is

\begin{aligned} \int _{\Omega }|u_{n}(t)|^q \le \frac{\mathrm{e}^{C\,t}}{[A C^{-1}\,(\mathrm{e}^{C\,\delta \,t} - 1)]^{\frac{1}{\delta }}} \qquad \text{ for } \text{ every }\, t \in (0,T). \end{aligned}

Hence, passing to the limit on n we deduce that, for almost every t in (0, T) we have

\begin{aligned} \int _{\Omega }|u(t)|^q \le \frac{\mathrm{e}^{C\,t}}{[A C^{-1}\,(\mathrm{e}^{C\,\delta \,t} - 1)]^{\frac{1}{\delta }}} = \frac{1}{(A C^{-1})^{\frac{1}{\delta }}}\frac{\mathrm{e}^{C\,t}}{(\mathrm{e}^{C\,\delta \,t} - 1)^{\frac{1}{\delta }}}. \end{aligned}
(2.18)

Recalling the value of A, we thus have

\begin{aligned} \int _{\Omega }\,|u(t)|^{q} \le C_4 \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}^{\frac{q(1-\theta )}{\delta \,\theta }}\,\frac{\mathrm {e}^{C\,t}}{(\mathrm {e}^{C\,\delta \,t} - 1)^{\frac{1}{\delta }}} \quad \text { for } \text { almost } \text { every }t \in (0,T), \end{aligned}

where $$C_4= \left( \frac{q{\mathcal {S}}C}{\alpha (q-1)}\right) ^{\frac{1}{\delta }}$$. Since

\begin{aligned} \delta = \frac{1}{\theta } - 1 = \frac{1-\theta }{\theta } = \frac{2}{N\,(q-1)}, \end{aligned}

we finally have that, for almost every t in (0, T)

\begin{aligned} \int _{\Omega }\,|u(t)|^{q} \le C_4 \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}^{q}\,\frac{\mathrm{e}^{C\,t}}{(\mathrm{e}^{C\,\delta \,t} - 1)^{\frac{1}{\delta }}}. \end{aligned}
(2.19)

Recalling that $$\mathrm{e}^{C\,\delta \,t} - 1 \ge C\,\delta \,t$$ for every t (since $$t \mapsto \mathrm{e}^{C\,\delta \,t}$$ is convex), we therefore have that

\begin{aligned} \Vert u(t)\Vert _{L^{q}(\Omega )} \le C_1 \frac{\Vert u_0\Vert _{L^{1}(\Omega )}}{t^{\frac{1}{\delta q}}} \qquad \text { for } \text { almost } \text { every }\, t\, \text { in }\, (0,T), \end{aligned}
(2.20)

where

\begin{aligned} c_1 = \left[ C_4 \frac{\mathrm{e}^{C\,T}}{(C\delta )^{\frac{1}{\delta }}\,}\right] ^{\frac{1}{q}} \end{aligned}
(2.21)

and hence (2.4) is proved.

To conclude the proof we observe that by (2.19) it follows that even if $$u_0 \in L^{1}(\Omega )$$ it results

\begin{aligned} u \in L^{\infty }(t_0,T;L^{q}(\Omega )) \quad \text{ for } \text{ every }\, t_0\, \text{ in }\, (0,T),\, \text{ and } \text{ for } \text{ every }\, 1< q < +\infty . \end{aligned}
(2.22)

Moreover, it results

\begin{aligned} \Vert u\Vert _{_{L^{\infty }(t_0,T;L^{q}(\Omega ))}} \le c_0\, \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}, \end{aligned}
(2.23)

where $$c_{0}$$ is a constant depending only on $$t_0$$, q, N, $$\alpha$$, T and B.

Finally, using the previous regularity in (2.12) we deduce that for every $$q \ge 2$$

\begin{aligned} |\nabla |u|^{\frac{q}{2}-1}u| \in L^2(t_0,T;L^{2}(\Omega )), \quad |\nabla u| \in L^2(t_0,T;L^{2}(\Omega )) \quad \text{ for } \text{ every }\, t_{0}\, \text{ in }\, (0,T). \end{aligned}

$$\square$$

### Behaviour for t large

We show here that our problem admits a global solution u (defined in all the set $$\Omega \times (0,+\infty )$$) and we study its behavior for t large. To this aim, we recall that by a global solution of (1.1) we mean a function u which solves (1.1) for every $$T>0$$.

### Theorem 2.2

Assume (1.3) and that (1.2) and (1.5) hold true in $$\Omega \times (0,+\infty )$$. Let u be the solution of (1.1) given by Theorem 2.1. Then u can be extended to a global solution defined in $$\Omega \times (0,+\infty )$$ (that we denote again u) satisfying

\begin{aligned}&u \in L^{\infty }(t_0,+\infty ;L^{q}(\Omega )) \text{ for } \text{ every }\, t_0 > 0,\, \text{ and } \text{ for } \text{ every }\, 1< q < +\infty , \end{aligned}
(2.24)
\begin{aligned}&\Vert u\Vert _{_{L^{\infty }(t_0,+\infty ;L^{q}(\Omega ))}} \le \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}, \end{aligned}
(2.25)

where $$c_0$$ is as in (2.3) a constant depending only on $$t_0$$, q, N, $$\alpha$$ and B, and such that for every $$q \ge 2$$ and $$t_0 >0$$ we have

\begin{aligned} |\nabla |u|^{\frac{q}{2}-1}u| \in L^2_{\mathrm{loc}}(t_0,+\infty ;L^{2}(\Omega )), \quad |\nabla u| \in L^2_{\mathrm{loc}}(t_0,+\infty ;L^{2}(\Omega )). \end{aligned}
(2.26)

Moreover, if for some $$t_0 >0$$ it results

\begin{aligned} |E(x,t)| \le b(t)\, |B(x)|, \quad \text{ where }\, b \in L^{2}(t_0,+\infty )\, \text{ and }\, B(x) \in L^{N}(\Omega ), \end{aligned}
(2.27)

or more generally, if

\begin{aligned} |E(x,t)| \in L^2(t_0,+\infty ;L^{r}(\Omega )), \quad \text{ for } \text{ some }\, r > 2, \end{aligned}
(2.28)

then

\begin{aligned} \lim _{t \rightarrow + \infty } \Vert u(t)\Vert _{_{L^{q}(\Omega )}} = 0 \qquad \text{ for } \text{ every }\, 1< q < +\infty . \end{aligned}
(2.29)

### Proof

Let $$T>0$$ be arbitrarily fixed and let u be the solution of (1.1) given by Theorem 2.1. Hence u is obtained as the almost everywhere limit in $$Q_T$$ of a sequence $$u_{n}$$ in $$L^{2}(0,T;W^{1,2}_{0}(\Omega )) \cap C([0,T];L^{2}(\Omega ))$$ of weak solutions of (2.5). Notice that each $$u_{n}$$ can be extended to a global solution of (2.5), that we denote again $$u_{n}$$. Hence $$u_{n}$$ belongs to $$L^{2}_{\mathrm{loc}}([0,+\infty );W^{1,2}_{0}(\Omega )) \cap C_{\mathrm{loc}}([0,+\infty );L^{2}(\Omega ))$$, and solves (2.5) in every set $$Q_{\texttt {T}}$$ for every $$\texttt {T}>0$$ arbitrarily fixed. We show now that we can extend the solution u of (1.1) in $$Q_T$$ to a global solution. To this aim let us denote with $$u_{n}^{(1)}$$ the subsequence of $$u_{n}$$ that converges almost everywhere in $$Q_{2T}$$ to a weak solution $$u^{(2)}$$ of (1.1) in $$Q_{2T}$$. By construction $$u = u^{(2)}$$ in $$Q_T$$. Now, let us denote with $$u_{n}^{(2)}$$ the subsequence of $$u_{n}^{(1)}$$ that converges almost everywhere in $$Q_{3T}$$ to a weak solution $$u^{(3)}$$ of (1.1) in $$Q_{3T}$$. By construction $$u^{(2)} = u^{(3)}$$ in $$Q_{2T}$$. Iterating this procedure, the function $$u(x,t) \equiv u^{(n)}(x,t)$$ in $$Q_{nT}$$ (for every integer n) is well defined and is a global solution of (1.1) which, by construction and thanks to Theorem 2.1, satisfies (2.24)–(2.26).

We show now that the global solution u constructed above, under further assumptions on E, satisfies the other estimates enounced in Theorem 2.2.

To this aim, proceeding as in the proof of (2.8), and integrating in time, we deduce that for every $$0<t_0<t_1<t_2$$

\begin{aligned} \begin{array}{l} \displaystyle \int _{\Omega }\,|u_{n}(t_2)|^{q} - \int _{\Omega }\,|u_{n}(t_1)|^{q} + \frac{q \alpha \,(q-1)}{2} \int _{t_1}^{t_2}\!\!\int _{\Omega }|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} \\ \displaystyle \qquad \le \frac{q(q-1)}{2 \alpha }\,\int _{t_1}^{t_2}\!\!\int _{\Omega }\,|E(x,t)|^{2}\,|u_{n}(t)|^{q}. \end{array} \end{aligned}
(2.30)

Now, thanks to the Poincaré inequality we deduce

\begin{aligned} \begin{array}{l} \displaystyle \int _{\Omega }\,|u_{n}(t_2)|^{q} - \int _{\Omega }\,|u_{n}(t_1)|^{q} + \frac{2 \alpha \,(q-1) C_P}{q} \int _{t_1}^{t_2}\!\!\int _{\Omega }| u_{n}(t)|^{q} \\ \displaystyle \qquad \le \frac{q(q-1)}{2 \alpha } \,\int _{t_1}^{t_2}\!\!\int _{\Omega }\,|E(x,t)|^{2}\,|u_{n}(t)|^{q}, \end{array} \end{aligned}
(2.31)

where $$C_P$$ is the Poincaré constant. We can rewrite the previous estimate in the following way

\begin{aligned} \int _{\Omega }\,|u_{n}(t_2)|^{q} - \int _{\Omega }\,|u_{n}(t_1)|^{q} + M \int _{t_1}^{t_2}\!\!\int _{\Omega }| u_{n}(t)|^{q} \le \int _{t_1}^{t_2} g_{n}(t) dt. \end{aligned}
(2.32)

where we have set

\begin{aligned} M = \frac{2 \alpha \,(q-1) C_P}{q} \quad \mathrm{and} \quad g_{n}(t) = \frac{q(q-1)}{2 \alpha } \int _{\Omega }|E(x,t)|^{2}\,|u_{n}(t)|^{q}. \end{aligned}

Applying Proposition 3.2 in  we obtain

\begin{aligned} \int _{\Omega }\,|u_{n}(t)|^{q} \le \int _{\Omega }\,|u_{n}(t_0)|^{q} \mathrm{e}^{-Mt} + \int _{t_0}^t g_{n}(s) ds, \end{aligned}

and if $$g_{n} \in L^1(t_0,+\infty )$$ (again by Proposition 3.2) we get for every $$t>t_0$$

\begin{aligned} \int _{\Omega }\,|u_{n}(t)|^{q} \le \Lambda _{n} \mathrm{e}^{-\frac{M}{2}t} + \int _{\frac{t}{2}}^t g_{n}(s) ds, \end{aligned}
(2.33)

where

\begin{aligned} \Lambda _{n} = \int _{\Omega }\,|u_{n}(t_0)|^{q} + \Vert g_{n}\Vert _{L^1(t_0,+\infty )}. \end{aligned}

Notice that thanks to (2.25) it results

\begin{aligned} \Lambda _{n} \le \Lambda _1 \equiv c_0^q \Vert u_0\Vert ^q_{L^1(\Omega )} + \Vert g_{n}\Vert _{L^1(t_0,+\infty )}. \end{aligned}
(2.34)

Suppose now that

\begin{aligned} |E(x,t)| \le b(t) |B(x)|, \quad b \in L^2(t_0,+\infty ). \end{aligned}
(2.35)

Then

\begin{aligned}&\Vert g_{n}\Vert _{_{L^1(t_0,+\infty )}} \le c \int _{t_0}^{+\infty }\!\!\int _{\Omega }b(t)^2|B(x)|^2|u_{n}|^q \\&\quad \le c \int _{t_0}^{+\infty }\!\!\int _{\Omega }b^2 \left[ |B(x)|^N + |u_{n}|^{\frac{Nq}{N-2}} \right] \\&\quad \le c \left( \int _{t_0}^{+\infty }b^2 \right) \left[ \int _{\Omega }|B|^N + \Vert u_{n}\Vert _{_{L^{\infty }(t_0,+\infty ;L^{\frac{Nq}{N-2}}(\Omega ))}}^{\frac{Nq}{N-2}}\right] \\&\quad \le C \left( \int _{t_0}^{+\infty }b^2 \right) \left[ \int _{\Omega }|B|^N + \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}^{\frac{Nq}{N-2}}\right] , \end{aligned}

and

\begin{aligned} \int _{\frac{t}{2}}^t g_{n}(s) ds \le C \bigg (\int _{\frac{t}{2}}^{t}b^2 \bigg ) \left[ \int _{\Omega }|B|^N + \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}^{\frac{Nq}{N-2}}\right] . \end{aligned}

If instead one assumes that

\begin{aligned} E \in L^2(t_0,+\infty ;L^{r}(\Omega )), \quad \text{ for } \text{ some }\, r > 2, \end{aligned}
(2.36)

then it results

\begin{aligned} \Vert g_{n}\Vert _{_{L^1(t_0,+\infty )}}\le & {} c \int _{t_0}^{+\infty }\!\!\int _{\Omega }|E(x,t)|^2|u_{n}|^q \le c \int _{t_0}^{+\infty }\!\! \left( \int _{\Omega }|E(x,t)|^{r}\right) ^{\frac{2}{r}}\left( \int _{\Omega }|u_{n}|^{\frac{q r}{r-2}} \right) ^{1- \frac{2}{r}} \nonumber \\\le & {} c \left[ \int _{t_0}^{+\infty }\!\! \left( \int _{\Omega }|E(x,t)|^{r}\right) ^{\frac{2}{r}}\right] \Vert u_{n}\Vert _{_{L^{\infty }(t_0,+\infty ;L^{\frac{q r}{r-2}}(\Omega ))}}^{q} \nonumber \\\le & {} C \left[ \int _{t_0}^{+\infty }\!\! \left( \int _{\Omega }|E(x,t)|^{r}\right) ^{\frac{2}{r}}\right] \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}^{q}, \end{aligned}

and

\begin{aligned} \int _{\frac{t}{2}}^t g_{n}(s) ds \le C \left[ \int _{\frac{t}{2}}^{t}\!\! \left( \int _{\Omega }|E(x,t)|^{r}\right) ^{\frac{2}{r}}\right] \Vert u_{0}\Vert _{_{L^{1}(\Omega )}}^{q}. \end{aligned}

Under either assumption (2.35) or (2.36) one therefore has that

\begin{aligned} \Vert g_{n}\Vert _{_{L^1(t_0,+\infty )}} \le C, \qquad \lim _{t \rightarrow +\infty }\,\int _{\frac{t}{2}}^{t}\,g_{n}(s)\,ds = 0, \quad \text{ uniformly } \text{ with } \text{ respect } \text{ to }\, n. \end{aligned}

Thus,

\begin{aligned} \lim _{t \rightarrow + \infty } \int _{\Omega }\,|u_{n}(t)|^{q} = 0 \qquad \hbox { uniformly with respect to}\ n, \end{aligned}
(2.37)

so that (recalling that the solution u of (1.1) is the limit of $$u_{n}$$)

\begin{aligned} \lim _{t \rightarrow + \infty } \int _{\Omega }\,|u(t)|^{q} = 0. \end{aligned}

$$\square$$

## $$L^{\infty }$$-regularity

In this section we prove the following result.

### Theorem 3.1

Assume (1.2), (1.3), (1.5) and

\begin{aligned} |E| \in L^s(0,T;L^r(\Omega )) \quad \text{ with }\, r > 2\, \text{ and } s \ge 1 \text{ such } \text{ that } \frac{N}{2r} + \frac{1}{s} < \frac{1}{2}. \end{aligned}
(3.1)

Then there exists a solution u of (1.1) satisfying

\begin{aligned} u \in L^{\infty }(\Omega \times (t_0,T)) \quad \text { for every } \,\, t_0 \in (0,T) .\end{aligned}
(3.2)

Moreover,

\begin{aligned} \Vert u(t)\Vert _{_{L^{\infty }(\Omega )}} \le c \frac{\Vert u_{0}\Vert _{_{L^{1}(\Omega )}}}{t^{\frac{N}{2}}}, \quad \text{ for } \text{ every } t \in (0,T), \end{aligned}
(3.3)

where c depends only on $$\alpha$$, $$\beta$$, N, T and E.

Finally, if (1.2) and (1.5) hold true in $$\Omega \times (0,+\infty )$$ then u can be extended to a global solution (that we denote again u) defined in all $$\Omega \times (0,+\infty )$$; this extension coincides with the one given by Theorem 2.2; if |E| belongs to $$L^s_{\mathrm{loc}}(0,+\infty ;L^{r}(\Omega ))$$ and (2.37) holds true (for example if (2.35) or (2.36) holds true) then it results

\begin{aligned} \lim _{t \rightarrow + \infty }\Vert u(t)\Vert _{_{L^{\infty }(\Omega )}} = 0. \end{aligned}
(3.4)

### Proof

We recall that under the assumptions of Theorem 2.1 (see Sect. 2.1) there exists a solution u of (1.1) satisfying the regularity property (2.2) and estimate (2.3). Hence, if we assume that

\begin{aligned} |E| \in L^s(0,T;L^{r}(\Omega ))) \quad \text{ with } r>2 \text{ and } s \ge 1 \text{ such } \text{ that } \displaystyle \frac{N}{2r} + \frac{1}{s} < \frac{1}{2}, \end{aligned}
(3.5)

then there exist $${\overline{s}} > 1$$ and $${\overline{r}}$$ (depending on r and s) such that it results

\begin{aligned} |F| {\mathop {=}\limits ^{\mathrm {def}}} |E(x,t)\,u| \in L^{{\overline{s}}}(t_{0},T;L^{{\overline{r}}}(\Omega ))) \quad \text { with } {\overline{r}}> 2 \text { such } \text { that } \frac{N}{2 {\overline{r}}} + \frac{1}{{\overline{s}}} < \frac{1}{2}, \text { for every } t_{0} > 0.\nonumber \\ \end{aligned}
(3.6)

Notice that (3.5) is satisfied if for example $$|E| \in L^{\infty }(0,T;L^{r}(\Omega ))$$ with $$r >N$$.

Thus, if we assume (3.5) we can apply Theorem 8.1 at page 192 of  to the solution u of (1.1) constructed in Theorem 2.1 to conclude that

\begin{aligned} u \in L^{\infty }(\Omega \times (t_0,T)) \quad \text{ for } \text{ every } t_0 \in (0,T). \end{aligned}
(3.7)

Moreover it is also possible to estimate the $$L^\infty$$-norm of u in dependence of the data and of the $$L^2$$-norm of u.

Now to complete the proof of the theorem we need to show the behavior of the $$L^\infty$$-norm of u near zero and for t large.

To this aim, we recall that applying Theorem 2 at page 18 of  we obtain the following estimate

\begin{aligned} \Vert u\Vert _{L^{\infty }(Q(\rho ))} \le C \left[ \rho ^{-\frac{N+2}{2}}\Vert u\Vert _{L^2(Q(3\rho ))} + \rho ^{\theta } \Vert |uE|\Vert _{{\overline{s}},{\overline{r}},Q(3 \rho )}\right] \end{aligned}
(3.8)

where C is a constant depending only on the data (i.e. on $$\alpha$$, $$\beta$$, $$\theta$$ and N) $$Q(\rho ) = R(\rho ) \times (t_0-\rho ^2,t_0)$$, $$R(\rho )$$ is the open cube in $$R^N$$ of edge lenght $$\rho$$ centered in $$x_0 \in \Omega$$, $$\rho >0$$ such that $$Q(3 \rho ) \subset Q_T$$, $$\Vert \cdot \Vert _{s_0,r_0,Q(3 \rho )}$$ denotes the norm $$\Vert \cdot \Vert _{ L^{s_0}(t_0-(3 \rho )^2,t_0;L^{r_0}(K(3\rho )))}$$ and $$\theta \in (0,1)$$ is defined as follows

\begin{aligned} \theta \le 1 - \left( \frac{N}{2{\overline{r}}} + \frac{1}{{\overline{s}}} \right) \quad \mathrm{and} \quad \theta \le 1 - \frac{1}{r}. \end{aligned}
(3.9)

We conclude the proof distinguishing the two cases of t near zero and t large.

Case 1: t near zero. We observe that it results

\begin{aligned} \Vert u\Vert _{L^2(Q(3\rho ))} \le 3 \rho \Vert u\Vert _{L^{\infty }(t_0-(3 \rho )^2,t_0;L^2(K(3\rho )))} \end{aligned}
(3.10)

which implies

\begin{aligned} \rho ^{-\frac{N+2}{2}}\Vert u\Vert _{L^2(Q(3\rho ))} \le 3 \rho ^{-\frac{N}{2}} \Vert u\Vert _{L^{\infty }(t_0-(3 \rho )^2,t_0;L^2(K(3\rho )))} \end{aligned}
(3.11)

Hence, choosing $$(3 \rho )^2 = \frac{t_0}{2}$$ and t arbitrarily in $$(\frac{17}{18}t_0,t_0)$$ (hence $$t = \theta _0 t_0$$ with $$\theta _0 \in (\frac{17}{18},1)$$) and using estimate (2.4) we deduce

\begin{aligned} \rho ^{-\frac{N+2}{2}}\Vert u\Vert _{L^2(Q(3\rho ))} \le C t^{-\frac{N}{4}} \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{4}}} = C \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}}}. \end{aligned}
(3.12)

Here with C or c we denote positive constants, depending only on the data in the structure conditions and on E, which can vary from line to line. Notice that with such a choice of $$\rho$$ the left hand side of (3.8) is greater then $$\Vert u(t)\Vert _{L^{\infty }(K(\rho ))}$$. Hence, it remains to estimate the last term in the right hand side of (3.8).

Let us consider the particular case when $$|E| \in L^{\infty }(0,T;L^r(\Omega ))$$ with $$r >N$$ (i.e. (3.5) with $$s=+\infty$$). Then $$|uE| \in L^{\infty }(t_0,T;L^{{\overline{r}}}(\Omega ))$$ for every $$N< {\overline{r}} < r$$ for every $$t_{0}>0$$, and we have

\begin{aligned} \Vert |uE|\Vert _{\infty ,{\overline{r}},Q(3 \rho )} \le \Vert |E|\Vert _{\infty ,r,Q(3 \rho )} \Vert u\Vert _{\infty ,\frac{{\overline{r}}r}{r-{\overline{r}}},Q(3\rho )} \end{aligned}
(3.13)

Hence, by (3.13) and (2.4) we deduce that

\begin{aligned} \rho ^{\theta } \Vert |uE|\Vert _{{\overline{s}},{\overline{r}},Q(3 \rho )} \le c t^{\frac{\theta }{2}} \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}\left( 1 - \frac{r-{\overline{r}}}{{\overline{r}}r}\right) }} \end{aligned}
(3.14)

which implies, since $$\theta >0$$ and being possible any choice of $${\overline{r}}$$ satisfying $$N< {\overline{r}} < r$$, the following estimate

\begin{aligned} \rho ^{\theta } \Vert |uE|\Vert _{{\overline{s}},{\overline{r}},Q(3 \rho )} \le c \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}}}. \end{aligned}
(3.15)

Now, by (3.15), (3.12) and (3.8) we can conclude that for t in (0, T) it results

\begin{aligned} \Vert u(t)\Vert _{L^{\infty }(K(\rho ))} \le c \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}}}, \end{aligned}

which implies, thanks to the arbitrariness of $$K(\rho )$$,

\begin{aligned} \Vert u(t)\Vert _{L^{\infty }(\Omega )} \le c \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}}}. \end{aligned}
(3.16)

We conclude the proof considering the general case when |E| satisfies (3.5) with $$s \not =+\infty$$. In this case we estimate the last term in (3.8) as follows

\begin{aligned}&\rho ^{\theta } \Vert |uE|\Vert _{{\overline{s}},{\overline{r}},Q(3 \rho )} \le \rho ^{\theta } \Vert |E|\Vert _{s,r,Q(3 \rho )}\Vert |u|\Vert _{\frac{{\overline{s}}s}{s-{\overline{s}}},\frac{{\overline{r}}r}{r-{\overline{r}}},Q(3 \rho )} \nonumber \\&\quad \le \rho ^{\theta + 2 \frac{s-{\overline{s}}}{{\overline{s}}s}}\Vert |E|\Vert _{s,r,Q(3 \rho )}\Vert |u|\Vert _{\infty ,\frac{{\overline{r}}r}{r-{\overline{r}}},Q(3 \rho )}, \end{aligned}

for every $${\overline{s}}<s$$ and $${\overline{r}} < r$$ satisfying (3.6). Hence, by the previous estimate and (2.4) we deduce, for every t in (0, T), that

\begin{aligned} \rho ^{\theta } \Vert |uE|\Vert _{{\overline{s}},{\overline{r}},Q(3 \rho )} \le c t^{\frac{\theta }{2}+\frac{s-{\overline{s}}}{{\overline{s}}s}} \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}\left( 1 - \frac{r-{\overline{r}}}{{\overline{r}}r}\right) }} \end{aligned}
(3.17)

which implies (thanks to the arbitrariness of $${\overline{s}}<s$$ and $${\overline{r}} < r$$) that (3.15) holds true and hence estimate (3.16) is true also in this more general case.

Case 2: t large. To study the solution for t large we need to assume further structure assumptions to guarantee the existence of a solution defined for every value of t. Hence, let us assume that (1.2) and (1.5) hold true in $$\Omega \times (0,+\infty )$$. Thus, by Theorem 2.2u can be extended to a global solution (that we denote again u) defined in all $$\Omega \times (0,+\infty )$$. Moreover, assuming also that $$|E| \in L^s_{loc}(0,+\infty ;L^r(\Omega ))$$ we obtain that estimate (3.8) holds true for every $$Q(3\rho ) \subset \Omega \times (0,+\infty )$$.

Hence, if (2.37) holds true (for example if (2.35) or (2.36) holds true) then by (3.8) we deduce that also

\begin{aligned} \lim _{t \rightarrow + \infty }\Vert u(t)\Vert _{L^{\infty }(K(\rho ))} = 0 \end{aligned}
(3.18)

and hence, thanks to the arbitrariness of $$K(\rho )$$, we conclude that

\begin{aligned} \lim _{t \rightarrow + \infty }\Vert u(t)\Vert _{L^{\infty }(\Omega )} = 0 \end{aligned}
(3.19)

$$\square$$

## E does not belong to $$L^{\infty }(0,T;L^{N}(\Omega ))$$

We consider here a particular case when $$E \not \in L^{\infty }(0,T;L^N(\Omega ))$$. We have the following result.

### Theorem 4.1

Assume that 0 belongs to $$\Omega$$, that (1.2), (1.3) hold, and that

\begin{aligned} |E(x,t)| \le \frac{\gamma }{|x|} \, \quad \mathrm{with} \quad \gamma < \alpha (N-2). \end{aligned}
(4.1)

Then there exists a solution u of (1.1) satisfying the following estimate for every $$1 \le q <(N-2)\frac{\alpha }{\gamma }$$ and for a.e. $$t \in (0,T)$$

\begin{aligned} \Vert u(t)\Vert _{L^q(\Omega )} \le c \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}\frac{1}{q'}}}, \end{aligned}
(4.2)

where $$c = c(N,q,\alpha ,\gamma )$$.

### Proof

Proceeding as in the proof of Theorem 2.1 we deduce that (2.8) holds true with $$B(x)=\frac{\gamma }{|x|}$$, i.e.

\begin{aligned} \frac{1}{q} \frac{d}{dt} \int _{\Omega }|u_{n}(t)|^{q} + \frac{\alpha \,(q-1)}{2} \int _{\Omega }|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} \le \frac{(q-1)}{2\alpha }\,\gamma ^2\int _{\Omega }\frac{|u_{n}(t)|^{q}}{|x|^2}. \end{aligned}

Recalling Hardy inequality, we thus have that

\begin{aligned} \begin{array}{l} \displaystyle \frac{1}{q} \frac{d}{dt} \int _{\Omega }|u_{n}(t)|^{q} + \frac{\alpha \,(q-1)}{2} \int _{\Omega }|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} \\ \displaystyle \quad \le \frac{(q-1)}{2\alpha }\frac{\gamma ^2}{{\mathcal {H}}^2}\int _{\Omega }|\nabla [|u_{n}(t)|^{\frac{q}{2}-1}u_{n}]|^2 =\frac{(q-1)}{2\alpha }\frac{\gamma ^2\,q^2}{4{\mathcal {H}}^2}\int _{\Omega }|\nabla u_{n}(t)|^2\,|u_{n}(t)|^{q-2}, \end{array} \end{aligned}

that is

\begin{aligned} \frac{1}{q} \frac{d}{dt} \int _{\Omega }|u_{n}(t)|^{q} + \frac{1}{2}\alpha (q-1)\bigg (1 -\frac{1}{\alpha ^2}\frac{\gamma ^2\,q^2}{4{\mathcal {H}}^2}\bigg ) \int _{\Omega }|\nabla u_{n}(t)|^{2}|u_{n}(t)|^{q-2} \le 0\nonumber \\ \end{aligned}
(4.3)

We observe that, since $${\mathcal {H}} = \frac{N-2}{2}$$, it results

\begin{aligned} \gamma \,q< 2 {\mathcal {H}}\,\alpha \quad \Longleftrightarrow \quad q<(N-2)\frac{\alpha }{\gamma }. \end{aligned}
(4.4)

Since we need to have $$q \ge 1$$, by (4.4) we deduce that under the following assumption

\begin{aligned} \gamma < \alpha (N-2) \end{aligned}
(4.5)

estimate (4.3) holds true for every $$1 \le q <(N-2)\frac{\alpha }{\gamma }$$. By Sobolev inequality and (4.3) it follows for every $$0<t_1 < t_2 \le T$$

\begin{aligned} \int _{\Omega }|u_{n}(t_2)|^{q} - \int _{\Omega }|u_{n}(t_1)|^{q} + c_1 \int _{t_1}^{t_2} \Vert u_{n}\Vert ^q_{\frac{2^{*}}{2}q} \le 0 \end{aligned}
(4.6)

where $$c_1 = \frac{1}{2}\alpha (q-1)\bigg (1 -\frac{1}{\alpha ^2}\frac{\gamma ^2\,q^2}{4{\mathcal {H}}^2}\bigg )\frac{4}{q^2S}$$. We recall that since $$E\in L^2(Q_T)$$ it results

\begin{aligned} \Vert u_{n}(t)\Vert _{L^1(\Omega )} \le \Vert u_0\Vert _{L^1(\Omega )} \end{aligned}

(see the proof of Lemma 3.1 in , or (2.1)). Hence, we can apply Theorem 2.1 in  (since the exponents in the integral inequality (4.6) satisfy all the needed requirements) and we can conclude that if (4.5) is satisfied then for every $$1 \le q <(N-2)\frac{\alpha }{\gamma }$$ it results

\begin{aligned} \Vert u_{n}(t)\Vert _{L^q(\Omega )} \le c_3 \frac{\Vert u_0\Vert _{L^1(\Omega )}}{t^{\frac{N}{2}\frac{1}{q'}}}, \end{aligned}
(4.7)

where $$c_3 = \left( \frac{N(q-1)}{2c_1}\right) ^{\frac{N}{2}\frac{1}{q'}}$$, from which the result follows (since the solution u of (1.1) is the limit of $$u_{n}$$). $$\square$$

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## Acknowledgements

The authors would like to thank Google and its Google Meet software for supporting all the discussions (both preliminary and final) which generated the present paper during the COVID-19 lockdown in Italy.

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